import numpy as np
import time

# 定義一個(gè)Python列表
python_list = [1, 2, 3, 4, 5]

# 使用純Python循環(huán)計(jì)算平方的時(shí)間測(cè)試
def measure_time(function, argument):
    start_time = time.time()
    result = function(argument)
    end_time = time.time()
    return result, end_time - start_time

# 定義純Python循環(huán)計(jì)算平方的函數(shù)
def square_elements_python(lst):
    squared_lst = []
    for num in lst:
        squared_lst.append(num ** 2)
    return squared_lst

# 計(jì)算并輸出純Python循環(huán)方法執(zhí)行時(shí)間和結(jié)果
python_squares, python_time = measure_time(square_elements_python, python_list)
print(f"純Python循環(huán)方法: {python_squares}, Time taken: {python_time} seconds")

# 轉(zhuǎn)換為NumPy數(shù)組并使用向量化操作
start_time = time.time()
numpy_array = np.array(python_list)
numpy_squares = numpy_array ** 2
end_time = time.time()

# 輸出NumPy向量化操作執(zhí)行時(shí)間
numpy_time = end_time - start_time
print(f"NumPy向量化操作: {numpy_squares.tolist()}, Time taken: {numpy_time} seconds")

純Python循環(huán)方法: [1, 4, 9, 16, 25], Time taken: 4.5299530029296875e-06 seconds
NumPy向量化操作: [1, 4, 9, 16, 25], Time taken: 0.00020122528076171875 seconds

import numpy as np
from numba import jit

@jit(nopython=True)
def sum_array(arr: np.ndarray) -> float:
    result = 0.0
    for i in range(arr.shape[0]):
        result += arr[i]
    return result

arr = np.array([1.0, 2.0, 3.0, 4.0, 5.0])
print(sum_array(arr))

# 避免：
list1 = [1, 2, 3]
list2 = [4, 5, 6]
new_list = list1 + list2

# 推薦：
list1 = [1, 2, 3]
list2 = [4, 5, 6]
# 這里不會(huì)創(chuàng)建新的列表對(duì)象，而是直接在原地?cái)U(kuò)展list1
list1.extend(list2)

numbers = [1, 2, 3, 4, 5]
squared_numbers = [num ** 2 for num in numbers]
for squared_num in squared_numbers:
    print(squared_num)

1
4
9
16
25

numbers = [1, 2, 3, 4, 5]
squared_numbers = (num ** 2 for num in numbers)
for squared_num in squared_numbers:
    print(squared_num)

import multiprocessing as mp

def cpu_bound_task(n):
    # 模擬的CPU密集型計(jì)算任務(wù)
    result = 0
    for i in range(n):
        result += i * i
    return result

if __name__ == "__main__":
    inputs = [1_000_000 + x for x in range(10)]  # 多個(gè)需要處理的數(shù)據(jù)單元

    with mp.Pool(processes=mp.cpu_count()) as pool:  # 使用所有可用CPU核心數(shù)
        results = pool.map(cpu_bound_task, inputs)  # 將任務(wù)分配到各個(gè)進(jìn)程中并行處理

    print(f"Results: {results}")

from functools import lru_cache

@lru_cache(maxsize=None)  # 緩存所有結(jié)果，可以根據(jù)實(shí)際情況設(shè)置緩存大小
def expensive_computation(x):
    # 假設(shè)這是一個(gè)計(jì)算成本很高的函數(shù)
    print("Computing...")
    return x ** x

# 第一次調(diào)用時(shí)會(huì)執(zhí)行計(jì)算
result1 = expensive_computation(5)
# 第二次調(diào)用時(shí)會(huì)從緩存中獲取結(jié)果，不再執(zhí)行計(jì)算
result2 = expensive_computation(5)
print(result1 == result2)

import asyncio

async def fetch(url):
    # 異步網(wǎng)絡(luò)請(qǐng)求
    response = await asyncio.get_event_loop().run_in_executor(None, fetch_from_network, url)
    return response

async def main():
    tasks = [fetch('http://example.com') for _ in range(10)]
    responses = await asyncio.gather(*tasks)
    for response in responses:
        process_response(response)

# 啟動(dòng)事件循環(huán)
asyncio.run(main())

# example_cython.pyx
def cython_power(int x):
    return x ** x

$ cython example_cython.pyx
$ gcc -shared -pthread -fPIC -fwrapv -O2 -Wall -fno-strict-aliasing -I/usr/include/python3.7 -o example_cython.cpython-37m-x86_64-linux-gnu.so example_cython.c

import example_cython

# 使用Cython優(yōu)化后的函數(shù)
result = example_cython.cython_power(5)
print(result)